 Welcome back to our lecture series, Math 42-20, abstract algebra 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Mistledine. In the contents of lecture 17, we want to revisit the idea of the dihedral group, that is the group Dn, which we had introduced previously as the symmetries of the regular n-gon, that is an n-sided polygon where all sides and angles are congruent to each other. We'd seen this previously, but much like the other permutation groups, particularly Sn, that we've seen in this chapter, we want to revisit this and simplify our approach, our understanding to the dihedral group using our more advanced knowledge of abstract algebra, if we can even call it that at this time. Be aware that in the past we've studied the groups D3 and D4 extensively. D3 was of course the symmetry group of the equilateral triangle, and D4 is the symmetry group for the square. In general, of course, any the Dn, for example, is going to be the symmetry group of a regular n-gon. Let's talk a little bit of what that would look like. How should we understand this thing? For any n-gon, I want you to be aware that this thing can be inscribed inside of a circle whose center coincides with the center of the n-gon here. Imagine you can see a circle. It's not the best drawn circle. I do apologize for that, but do imagine we have a circle that's drawn, and we're going to have the x-axis and y-axis meet at the center of this circle right here. What we're going to do is every regular n-gon can be inscribed inside of a circle. We're going to choose not just any old circle, we're going to choose the unit circle. Great, we love the unit circle, but we're going to choose the unit circle in the complex plane. What we're going to do is we're going to associate the vertices of our regular n-gon with complex roots of unity. We're always going to position in the following way that since one, the number one, is always a root of unity, we're going to make sure one is a vertex for our regular n-gon. Then we would take the next vertices to be the principal root of unity, n-th root of unity, and then we just continue around the circle taking each and every one of the n-th roots of unity until we have them all labeled. In this example, I'll just draw the sixth roots of unity. We're going to have, for example, one. We're going to have the element which we're going to call zeta-6, which be where this is going to be e to the 2pi, 2pi i over 6, or if you prefer, e to the pi i-thirds. Then the next one over here, you would get zeta-squared, which would coincide with the next one here. You're going to get e. We're just going to double the angle now, so you're going to get 2pi i-thirds. Then the next one zeta cube, that's just the number negative one. Then the next one zeta to the fourth, this would be e to the 2pi, I guess it would be, excuse me, it would be 4pi i over 3. Then the next one zeta 5, this would be e to the 5pi i over 3. Then, of course, you could get back to one, which is the same thing as zeta-6 because these, of course, are all sixth roots of unity. What we can do is we can identify the regular in-gone with this in-gone in the complex plane. That is, we're going to identify the vertices of our polygon with these complex roots of unity. No, it's not a perfect, it's not by any means a regular hexagon here drawn in the plane, but you'll give me some leniency here. This is the best I can draw for the moment. Of course, it always contains the number one. You're going to have your principal roots of unity, which would include zeta and zeta to the 5 right here. You're also going to have, since it's the sixth roots, you'll have your third roots of unity right here, which would be zeta 2 and zeta to the fourth. Then you'll have your second root of unity, which is negative 1. All of those will be the vertices of our polygon here. Now, we're going to define two specific symmetries of our in-gone, and this notation we're going to use for every single regular in-gone. Hence, these will be elements of every single dihedral group. The first one is going to be based upon rotation. If we're thinking of the symmetries of the regular in-gone, then much like we saw in the previous slide associated to the sixth roots of unity, we can take the angle theta to the 2 pi over n. What we're going to do is we're going to define a map. We're going to call this map r. r will be rotation around the origin in the complex plane by the angle measure 2 pi over n, and we're going to go counter-clockwise. With regard to our regular in-gone, what this will do is this will take the the vertices of the 1, which of course is on the x-axis, and if you rotate it by one little tick, because 2 pi n will be the smallest angle that moves 1 to the principal in root of unity. This will of course have the effect that everyone gets moved over. If this is the first point on our in-gone, then the second point will go to 3. The 3 will go to 4. The 4 goes to 5. The 5 goes to 1, like you can see over here. Be aware that this first pentagon is what the original labeling is, and then as everyone moves by one, if you rotate it by a single click, and to be precise that would be a 2 pi in counter-clockwise rotation. We're going to call that 1 r. The other one we're going to introduce here is we're going to introduce a reflection map, which we're going to call s, for which s is going to be reflection across the real axis, or which is identified with the x-axis in the complex plane. Therefore, if you take, for example, points 2 and 5 on this regular in-gone, they will swap locations, so 2 becomes 5 and 5 becomes 2. Likewise, 3 and 4 on this pentagon will swap locations, and so 3 becomes 4 and 4 becomes 3. I want to mention that geometrically, the reason we chose this sift tree is somewhat significant, because if we think of the vertices of our in-gone as complex roots of unity, then reflection across the horizontal axis, which you'll always have a horizontal axis of reflection, whether you have an even number or odd number of vertices, this line here will always pass through one of the vertices the way we oriented, the vertex that corresponds to the number 1. If it's an odd number, then this will go through the midpoint of the opposite side. On the other hand, if it's an even number, then this line, of course, will go through with different vertex over here, which will necessarily correspond to negative 1, and is halfway through all of the roots of unity there. Geometrically, this reflection across the real axis will be complex conjugation. Again, there's this geometric and algebraic significance of choosing this as our canonical reflection. It's going to be complex conjugation. Then, of course, the rotation we mentioned earlier, this is the rotation that sends the number 1 to the principle in throat of unity. Again, there's some significance of choosing these canonical representations, R and S. They're kind of a big deal. We should mention that, of course, reflection is going to be an order 2 operation here, order 2 symmetry. If you reflect twice, you get back the original element. On the other hand, R, so we see that we see that S will be a map, which is order 2. R, on the other hand, it's a rotation. And because we chose sort of this minimal rotation 2 pi over n, we're going to see that this is going to be a rotation of order n. In particular, the cyclic subgroup generated by R is going to form the cyclic subgroup of order n. This will give you the rotational subgroup inside of the dihedral group. In fact, using notation we introduced previously, we can identify this rotational subgroup with Zn, the group of complex roots of unity, which is a cyclic group. This rotational group here can be identified with those groups of complex numbers here, where basically the angle theta of rotation can be identified with an in throat of unity in the complex plane. Now, why do we care about these two maps, these two symmetry specifically, R and S? Well, the thing is these two maps are going to generate the entire dihedral group. That is, we can say something like the dihedral group Dn is equal to the group R, S. So notice here, if we just have a single element, we talk about the cyclic subgroup generated by R. What that meant was, of course, the smallest subgroup of the group that contains R. So what we mean here is that R and S, we want the smallest subgroup that contains both R and S. And in the dihedral group, that's going to be the whole group. And we say that these things generate the group. Essentially, we're arguing that every symmetry in the dihedral group can be written as a product of R's, S's, and their inverses. And by comparison, of course, we proved in the previous lecture that Sn is generated by the set of transpositions. We show that every permutation can be written as a product of transposition. So the transpositions form a generating set for the whole group. And so that gives you a little bit of a preview of these maps R and S. And the next video will prove specifically this fact right here. So stay tuned. You should hopefully be able to see the link on the screen right now. Take a look at it if you would like.